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1. need help on gas questions
a uniform, narrow-bored tube is closed at one end, contains some dry air which is sealed by a thread of mercury 15.0cm long.

when the tube is vertical, with the closed end at the bottom,the air column is 20cm long, but when held horizontally, the air column is 24cm long.

calculate the atmospheric pressure.

thx
2. Re: need help on gas questions
Assuming temperature of the gas stays constant:
p1V1 = p2V2

Or in the case of a cylinder of constant area:
p1L1 = p2L2

Now, the mercury has the effect of increasing the pressure on the cylinder of air when it is vertical:
p2 = p1 + ρgLmercury

where ρ = the density of mercury

So subbing this in to the first equation:
p1L1 = (p1 + ρgLmercury)L2
p1(L1-L2) = (ρgLmercury)L2
p1 = (ρgLmercury)L2/(L1-L2)
p1 = (ρgLmercury)/([L1/L2]-1)

Sub in the numbers (I think the density of mercury is 5430 kgm-3)

p1 = (5430 kgm-3 x 9.81ms-2 x 0.15m)/(0.24m/0.20m - 1)
= 39 951 Nm-2
= 40 kPa or about 0.4 atm
Last edited by Morbo; 14-01-2007 at 17:56.
3. Re: need help on gas questions
Er - mercury is a bit denser than that. Often atmospheric pressure is caluclated in cm of Hg so if you work in units of pressure of cmHg then the problem is easy.

P x 24 = ( P + 15 ) x 20

4 P = 300

P = 75 cm Hg

(normal atm is 76cm hg )
4. Re: need help on gas questions
Crikey. Google failed me. It gave me the density of the planet Mercury

So my answer is (13 500/5430) times bigger = 2.5 * 40 = 100kPa
5. Re: need help on gas questions

Nice one
6. Re: need help on gas questions
teachercol im in love with you for solving this :P thanks